Given:ABCD is a rhombus.
To prove:DE congruent to BE.
In rombus, we know opposite angle are equal.
so, angle DCB = angle BAD
SINCE, ANGLE DCB= BAD
SO, In triangle DCA
angle DCA=angle DAC
similarly, In triangle ABC
angle BAC=angle BCA
since angle BCD=angle BAD
Therefore, angle DAC =angle CAB
so, opposite sides of equal angle are always equal.
so,sides DC=BC
Now, In triangle DEC and in triangle BEC
1. .DC=BC (from above)............(S)
2ANGLE CED=ANGLE CEB (DC=BC)....(A)
3.CE=CE (common sides)(S)
Therefore,DE is congruent to BE (from S.A.S axiom)
Answer:
Step-by-step explanation:
(6,18)
Answer:
B!
Step-by-step explanation:
Answer:7
Step-by-step explanation:
6p-4+7p+3=90
13p-1=90
13p=91
p=7
Answer:
The wrt=itten expressions are too difficult to interpret properly. I did my best but I don't see any of the answers as equivalent to <u>-3x - 2y, so please check my interpretations of the answer options.</u>
Step-by-step explanation:
"Negative 3 x minus one-half 4 y"
-3x- (1/2)4y
<u>or -3x - 2y</u>
<u>=====</u>
The answer options are too garbled for me to make sense of them. Are they:
1. 2 y minus 5 x minus one-half 2 y : 2y -5x - (1/2)2y; <u>y -5x</u> ?
2. 2 x Negative 2 x minus one-half 6 y : 2(-2x)- (1/2)6y; <u>-4x - 3y</u> ?
3. 3 x Negative 3 x minus three-fourths 4 y : 3x(-3x) - (3/4)4y; <u>-9x -3y</u> ?
4. one-fourth Negative 3 y minus three-fourths 7 y one-fourth minus 3 x:
(1/4)(-3y) - (3/4)7y - (1/4)(-3x); -(3/4)y -(21/4)y + (3/4)x; -(24/4)y + (3/4)x; ?
I don't see any of the answers as equivalent to <u>-3x - 2y, so please check my interpretations of the answer options.</u>
